How is the integral calculator finding this ∫ 0 a a 2 − x 2 ( a 2 + x 2 ) 2 d x while I get an undefined expression?

Question

How is the integral calculator finding this        ∫          0              a                          a        2            −              x        2                    (              a        2            +              x        2                    )        2              d  x while I get an undefined expression?

conveneau71 · Accepted Answer

Alternatively, both cloudy&#039;s suggestion and the following method (the third step uses   θ  =  arcsinh  ⁡      x    a   and integration by parts) also avoid having to divide by 0:                                  ∫                                            a              2                        −                          x              2                                            (                          a              2                        +                          x              2                                      )              2                                                        d                x                            =                            ∫                              2                          a              2                                            (                          a              2                        +                          x              2                                      )              2                                      −                  1                                    a              2                        +                          x              2                                                        d                x                            =                              2          a                ∫                              sech            3                    ⁡                      (            arcsinh            ⁡                          x              a                        )                                            d                          (          arcsinh          ⁡                      x            a                    )                −                  1          a                arctan        ⁡                  x          a                                    =                              1          a                          (          sech          ⁡          θ          tanh          ⁡          θ          +          ∫          sech          ⁡          θ                                d                    θ          )                −                  1          a                arctan        ⁡                  x          a                                    =                              1          a                          (          sinh          ⁡          θ                      sech            2                    ⁡          θ          +          arctan          ⁡                      x            a                    )                −                  1          a                arctan        ⁡                  x          a                +        C                            =                              1          a                          (                      x            a                    ×                      1                          1              +                                                (                                      x                    a                                    )                                2                                              )                +        C                            =                              x                                    a              2                        +                          x              2                                      +        C        .            Alternatively, use   θ  =  arctan  ⁡      x    a   or integrate by reduction formula.

How is the integral calculator finding this <msubsup> ∫ <mrow class="MJ

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